By Titu Andreescu

This demanding challenge e-book by means of popular US Olympiad coaches, arithmetic academics, and researchers develops a large number of problem-solving talents had to excel in mathematical contests and in mathematical examine in quantity concept. providing notion and highbrow pride, the issues in the course of the booklet motivate scholars to specific their rules in writing to provide an explanation for how they conceive difficulties, what conjectures they make, and what conclusions they achieve. using particular concepts and methods, readers will gather an exceptional realizing of the basic suggestions and ideas of quantity conception.

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**Extra info for 104 number theory problems: from the training of the USA IMO team**

**Example text**

We can repeat the above process of checking the rightmost digits of a2 and eliminate the rightmost even digits of a2 , if there is such a digit among the n rightmost digits of a2 . This process can be repeated at most n − 1 times because the units digit of a0 is 5. Thus, we can obtain a number ak , for some nonnegative integer k, such that ak is a multiple of 5n with its n rightmost digits all odd. Then bk = ak − (ak )·10n is a number that satisﬁes the conditions of the problem. We can replace the condition of odd digits by any collection of 5 digits that forms a complete set of residue classes modulo 5.

For x = b + 8i, where 0 ≤ i ≤ a − 1, 8(y + z) = 8(a − i) or y + z = a − i, which admits a − i − 1 ordered pairs (y, z) of positive integer solutions, namely, (1, a − i − 1), . . , (a − i − 1, 1). Hence there are a−1 a−1 (a − i − 1) = i=0 i= i=0 a(a − 1) 2 ordered triples satisfying the conditions of the problem. Solving a(a−1) = 666 2 gives a = 37. Therefore, the maximum value for n is equal to 37 · 8 + 7 = 303, obtained by setting b = 7. 41. Let b be an integer greater than 1. For any integer n ≥ 1 there is a unique system (k, a0 , a1 , .

Hence Sm is not a complete set of residue classes. Similarly, we can show the following result. 24. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer. Assume that S is a complete set of residue classes modulo m. The set T = aS + b = {as + b | s ∈ S} is also a complete set of residue classes modulo n. 26 104 Number Theory Problems Now we are better equipped to discuss linear congruence equations a bit further. 25. Let m be a positive integer. Let a be an integer relatively prime to m, and let b be an integer.